Why Doesn't This Series Converge?

Solution 1:

I don't think the series diverges.

$$ \frac{1 \cdot 3 \cdot 7 \cdots (2n-3)}{2 \cdot 4 \cdot 6 \cdots 2n}$$

$$ = \frac{{2n \choose n}}{(2n-1)4^n} = \frac{1}{(2n-1)\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$

using the approximation

$$\frac{ {2n \choose n}}{4^n} = \frac{1}{\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$

and so the series must converge! Perhaps you have a mistake in your computation?

For a more elementary proof of convergence, see the end of the answer.

I believe you should be able to compute it using the series expansion for

$$\frac{1}{\sqrt{1-x^2}} = \sum_{k=0}^{\infty} \frac{{2k \choose k} x^{2k}}{4^k}$$

(you will need to subtract some terms, divide by $x^2$ and integrate).

An elementary proof of convergence.

We will show that

if $\displaystyle S_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$ then $\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$

This proves the partial term of your series, which is $\displaystyle \sum_{k=2}^{n} \frac{S_k}{2k-1} < \sum_{k=2}^{\infty} \frac{1}{(2k-1)\sqrt{k+1}} < C$ (for some constant $C$), and thus is bounded above. Since the series is monotonically increasing and bounded above, it is convergent.

We will prove that $\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$ by induction on $n$.

For $n=1$ it is clearly true.

Now $S_{n+1} = S_n \frac{2n+1}{2n+2}$

Consider $\displaystyle 1 - \frac{2n+1}{2n+2} = \frac{1}{2n+2} \ge \frac{1}{\sqrt{n+2}(\sqrt{n+2} + \sqrt{n+1})} = \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+2}}$

Thus $\displaystyle \frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ and so

$\displaystyle S_{n} \le \frac{1}{\sqrt{n+1}} \Rightarrow S_{n+1} \le \frac{1}{\sqrt{n+2}}$

Solution 2:

You can use Taylor approximations here. Note that the ratio between consecutive terms is ${2n - 3 \over 2n} = \exp(\ln(1 - 3/2n)) = \exp(-{3 \over 2n} + O(1/n^2))$. So the product is comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$, which in turn is comparable to $\exp(-{3 \over 2} \ln(n))$ or $n^{-{3 \over 2}}$. Thus the series converges.

Comment: Seeing others computing the actual value... once you know it converges (absolutely) you can use the original power series to get the actual value. For $|x| < 1$, $(1 - x)^{1 \over 2} = 1 - {x \over 2} - {1 \over 2} \sum_{n=2}^{\infty} A_nx^n$, where $A_n$ is the $n$ term of the sum. Since all terms of the sum are positive and it converges for $x = 1$, if you take limits as $x$ goes to 1 you get the sum $S$ we're looking at. In other words, 0 = 1 - 1/2 - ${S \over 2}$ or $S = 1$.

And actually if you think about it.. if the series did diverge, taking limits of the series as $x$ goes to 1 from below would give infinity since all terms are positive. Thus just taking limits of both sides as $x$ goes to 1 gives both convergence and the correct value.